# Find range of $a$ if $\{x_n\}$ converges where $x_{n+1}=a^{x_{n}},n=1,2,\ldots$ and $x_1=a$ [duplicate]

For $$a>1, x_{n+1}=a^{x_{n}}, n=1,2, \ldots$$ and $$x_{1}=a$$. If $$\{x_{n}\}$$ converges, find range of $$a$$.

\begin{aligned} x_{n+1}-x_{n}&=f\left(x_{n}\right)-f\left(x_{n-1}\right)\\ &=f^{\prime}(\xi_n)\left(x_{n}-x_{n-1}\right)=\ln a e^{\xi_n \ln a}\left(x_{n}-x_{n-1}\right)\\ &=(\ln a)^2 e^{ \ln a(\xi_n+\xi_{n-1})}\left(x_{n-1}-x_{n-2}\right)\\ &\cdots\\ &=e^{(n-1)\ln\ln a+\ln a(\sum_{i=2}^{n}\xi_i)}(x_2-x_1) \end{aligned}

If $$\{x_n\}$$ converges, $$e^{(n-1)\ln\ln a+\ln a(\sum_{i=2}^{n}\xi_i)}<1 \implies (n-1) \ln \ln a<-\ln a\left(\sum_{i=2}^{n} \xi_{i}\right)$$

But I can't solve this inequation.

The answer is $$a \in (1,e^{\frac{1}{e}}]$$.

## marked as duplicate by Simply Beautiful Art, Adrian Keister, Paul Frost, José Carlos Santos limits StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 11 at 0:09

Let's say that $$y = x_\infty = a^{a^{a^{a^{...}}}}$$. You want to find the interval in $$a$$ for which $$y$$ converges. Then $$y = a^{y}$$ We then have that $$a = y^{\frac{1}{y}}$$
The maximum of this is $$e^{\frac{1}{e}}$$. This can be seen by looking at $$\frac{da}{dy}$$: $$\frac{da}{dy} = y^{\frac{1}{y}}\left(\frac{1-\ln(y)}{y^2}\right)$$
Setting this equal to $$0$$, we have that $$1-\ln(y) = 0 \rightarrow y = e \rightarrow a = e^{\frac{1}{e}}$$
Therefore, $$x_n$$ converges for $$a \in (1,e^{\frac{1}{e}}]$$ if $$a > 1$$.
• This does not prove convergence for $a \leq e^{1/e}$. – Kabo Murphy Aug 2 at 23:45
• @KaviRamaMurthy Why does it not? I'm showing that the inverse is only defined if $a < e^{\frac{1}{e}}$. – automaticallyGenerated Aug 3 at 0:06
• You start with the asumption that $x_n$ converges to some $y$. There is no $y$ to start with . What you have proved is that if $x_n$ converges then we must have $a \leq e^{1/e}$. You have to prove the converse of this also. – Kabo Murphy Aug 3 at 0:14